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          <h1 class="post-title" itemprop="name headline">PAT B1045/A1101</h1>
        

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        <p>There is a classical process named partition in the famous quick sort algorithm. In this process we typically choose one element as the pivot. Then the elements less than the pivot are moved to its left and those larger than the pivot to its right. Given N distinct positive integers after a run of partition, could you tell how many elements could be the selected pivot for this partition?</p>
<p>For example, given N = 5 and the numbers 1, 3, 2, 4, and 5. We have:</p>
<p>1 could be the pivot since there is no element to its left and all the elements to its right are larger than it;<br>3 must not be the pivot since although all the elements to its left are smaller, the number 2 to its right is less than it as well;<br>2 must not be the pivot since although all the elements to its right are larger, the number 3 to its left is larger than it as well;<br>and for the similar reason, 4 and 5 could also be the pivot.<br>Hence in total there are 3 pivot candidates.</p>
<p>Input Specification:</p>
<p>Each input file contains one test case. For each case, the first line gives a positive integer N (&lt;= 105). Then the next line contains N distinct positive integers no larger than 109. The numbers in a line are separated by spaces.</p>
<p>Output Specification:</p>
<p>For each test case, output in the first line the number of pivot candidates. Then in the next line print these candidates in increasing order. There must be exactly 1 space between two adjacent numbers, and no extra space at the end of each line.</p>
<p>Sample Input:<br>5<br>1 3 2 4 5<br>Sample Output:<br>3<br>1 4 5</p>
<p>著名的快速排序算法里有一个经典的划分过程：我们通常采用某种方法取一个元素作为主元，通过交换，把比主元小的元素放到它的左边，比主元大的元素放到它的右边。 给定划分后的N个互不相同的正整数的排列，请问有多少个元素可能是划分前选取的主元？</p>
<p>例如给定N = 5, 排列是1、3、2、4、5。则：</p>
<p>1的左边没有元素，右边的元素都比它大，所以它可能是主元；<br>尽管3的左边元素都比它小，但是它右边的2它小，所以它不能是主元；<br>尽管2的右边元素都比它大，但其左边的3比它大，所以它不能是主元；<br>类似原因，4和5都可能是主元。<br>因此，有3个元素可能是主元。</p>
<p>输入格式：</p>
<p>输入在第1行中给出一个正整数N（&lt;= 105）； 第2行是空格分隔的N个不同的正整数，每个数不超过109。</p>
<p>输出格式：</p>
<p>在第1行中输出有可能是主元的元素个数；在第2行中按递增顺序输出这些元素，其间以1个空格分隔，行末不得有多余空格。</p>
<p>输入样例：<br>5<br>1 3 2 4 5<br>输出样例：<br>3<br>1 4 5<br><figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br></pre></td><td class="code"><pre><span class="line">#include &quot;stdio.h&quot;</span><br><span class="line">#include &quot;algorithm&quot;</span><br><span class="line">#include &quot;string.h&quot;</span><br><span class="line">using namespace std;</span><br><span class="line">const int MAXN = 100010;</span><br><span class="line">const int INF = 0x7fffffff;</span><br><span class="line">int a[MAXN], leftMax[MAXN], rightMin[MAXN];</span><br><span class="line">int ans[MAXN], num = 0;</span><br><span class="line">int main()&#123;</span><br><span class="line">	int n;</span><br><span class="line">	scanf(&quot;%d&quot;, &amp;n);</span><br><span class="line">	for (int i = 0 ; i &lt; n ; i++ ) &#123;</span><br><span class="line">		scanf(&quot;%d&quot;, &amp;a[i]);</span><br><span class="line">	&#125;</span><br><span class="line">	leftMax[0] = 0;</span><br><span class="line">	for (int i = 1 ; i &lt; n ; i++) &#123;</span><br><span class="line">		leftMax[i] = max(leftMax[i - 1], a[i - 1]);</span><br><span class="line">	&#125;</span><br><span class="line">	rightMin[n - 1] = INF;</span><br><span class="line">	for (int i = n - 2 ; i &gt;=0 ; i--) &#123;</span><br><span class="line">		rightMin[i] = min(rightMin[i + 1],a[i + 1]);</span><br><span class="line">	&#125;</span><br><span class="line">	for (int i = 0; i &lt; n ;i++) &#123;</span><br><span class="line">		if(leftMax[i] &lt; a[i] &amp;&amp; a[i] &lt; rightMin[i])&#123;</span><br><span class="line">			ans[num++] = a[i];</span><br><span class="line">		&#125;</span><br><span class="line">	&#125;</span><br><span class="line">	printf(&quot;%d\n&quot;,num);</span><br><span class="line">	for (int i = 0; i &lt; num;i++ )&#123;</span><br><span class="line">		printf(&quot;%d&quot;, ans[i]);</span><br><span class="line">		if(i &lt; num -1)printf(&quot; &quot;);</span><br><span class="line">	&#125; </span><br><span class="line">	printf(&quot;\n&quot;);</span><br><span class="line">	return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure></p>

      
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